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Force of parallel plate capacitor

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data
    A parallel-plate capacitor has a plate area of .3m^2 and a plate separation of .1mm. If the charge on ech plate has a magnitude of 5*10^-6C then the force exerted by one plate on the other has what magnitude

    2. Relevant equations
    Q=E Ԑ A
    F=qE

    3. The attempt at a solution
    so to solve for E i used the first equation: 5*10^-6C=E*(10^-11)(.2) and got E=1.67*10^6 V/m
    First i want to make sure that i am moving in the right direction and also do i use 5*10^-6C as my q in the second equation?
     
  2. jcsd
  3. Mar 9, 2016 #2
    U= (Q^2)/(2c) = (Q^2)/(2ԐA) d and F=(dU)/(dd) = (-Q^2)/ (2ԐA)?
    so F= (5*10^-6C)^2/(2 * 10^-11 *.3)?

    is the answer then 4N?
     
  4. Mar 9, 2016 #3

    gneill

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    Staff: Mentor

    Your approximation for the value of ##\epsilon_o## is very crude. A good value would be ##\epsilon_o = 8.854 \times 10^{-12}~ F/m##, but even ##9 \times 10^{-12}~F/m## would be better.

    The appropriate approach to this problem is to find the electric field due to a single charged plate and then determine the force that it exerts on the charge of the other plate. Note that between the plates of a parallel plate capacitor the net field is due to the contribution of the fields from both plates; each plate contributes half the total field.
    What does the "(.2)" value represent?
     
  5. Mar 9, 2016 #4
    that was a typo i meant to write .3 m^2. ok i used the 8.85*10^-12 and got 4.71N is this more accurate
     
  6. Mar 9, 2016 #5

    gneill

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    Staff: Mentor

    That would be a good answer :smile:
     
  7. Mar 9, 2016 #6
    thank you very much!
     
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